An ion with mass number 37 possesses one unit | vedclass

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(A) Let the number of electrons in the ion be $x$.Since the ion has one unit of negative charge,the number of electrons is one more than the number of

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$_{17}^{37}Cl^{-}$

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(A) Let the number of electrons in the ion be $x$.Since the ion has one unit of negative charge,the number of electrons is one more than the number of protons $(Z)$.So,$Z = x - 1$.Number of neutrons $(n)$ $= x + 0.111 \, x = 1.111 \, x$.Mass number $(A)$ $= Z + n = 37$.Substituting the values: $(x - 1) + 1.111 \, x = 37$.$2.111 \, x = 38$.$x \approx 18$.Therefore,$Z = 18 - 1 = 17$.The element with atomic number $17$ is Chlorine $(Cl)$.Thus,the symbol of the ion is $_{17}^{37}Cl^{-}$.

An ion with mass number $37$ possesses one unit of negative charge. If the ion contains $11.1 \%$ more neutrons than the electrons,find the symbol of the ion.

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